Mixing
Differentiability through paths
$begingroup$
Let $f:mathbb{R}^{2} to mathbb{R}$ defined by
$$f(x) = begin{cases}displaystylefrac{x^{3}}{x^{2}+y^{2}},& (x,y)neq (0,0)\ 0,& (x,y) = (0,0)end{cases}.$$
Show that $f$ is not differentiable in $(0,0)$, however, show that for every differentiable path $lambda: (0,1) to mathbb{R}^{2}$, passing through origin, $f circ lambda$ is differentiable.
My problem is in the second part. If $lambda: mathbb{R} to mathbb{R}^{2}$ is a line pararallel to $e_{i}$ passing through $0$,
$lambda(t) = (0,0) + te_{i}$, I can show that $(fcirc lambda)'(0) = frac{partial}{partial x_{i}}f(0,0)$. I've tried to generalize this to an arbitrary path with domain $(0,1)$, but I couldnt. Can someone help me?
real-analysis derivatives
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}^{2} to mathbb{R}$ defined by
$$f(x) = begin{cases}displaystylefrac{x^{3}}{x^{2}+y^{2}},& (x,y)neq (0,0)\ 0,& (x,y) = (0,0)end{cases}.$$
Show that $f$ is not differentiable in $(0,0)$, however, show that for every differentiable path $lambda: (0,1) to mathbb{R}^{2}$, passing through origin, $f circ lambda$ is differentiable.
My problem is in the second part. If $lambda: mathbb{R} to mathbb{R}^{2}$ is a line pararallel to $e_{i}$ passing through $0$,
$lambda(t) = (0,0) + te_{i}$, I can show that $(fcirc lambda)'(0) = frac{partial}{partial x_{i}}f(0,0)$. I've tried to generalize this to an arbitrary path with domain $(0,1)$, but I couldnt. Can someone help me?
real-analysis derivatives
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
$endgroup$
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16
add a comment |
$begingroup$
Let $f:mathbb{R}^{2} to mathbb{R}$ defined by
$$f(x) = begin{cases}displaystylefrac{x^{3}}{x^{2}+y^{2}},& (x,y)neq (0,0)\ 0,& (x,y) = (0,0)end{cases}.$$
Show that $f$ is not differentiable in $(0,0)$, however, show that for every differentiable path $lambda: (0,1) to mathbb{R}^{2}$, passing through origin, $f circ lambda$ is differentiable.
My problem is in the second part. If $lambda: mathbb{R} to mathbb{R}^{2}$ is a line pararallel to $e_{i}$ passing through $0$,
$lambda(t) = (0,0) + te_{i}$, I can show that $(fcirc lambda)'(0) = frac{partial}{partial x_{i}}f(0,0)$. I've tried to generalize this to an arbitrary path with domain $(0,1)$, but I couldnt. Can someone help me?
real-analysis derivatives
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
$endgroup$
Let $f:mathbb{R}^{2} to mathbb{R}$ defined by
$$f(x) = begin{cases}displaystylefrac{x^{3}}{x^{2}+y^{2}},& (x,y)neq (0,0)\ 0,& (x,y) = (0,0)end{cases}.$$
Show that $f$ is not differentiable in $(0,0)$, however, show that for every differentiable path $lambda: (0,1) to mathbb{R}^{2}$, passing through origin, $f circ lambda$ is differentiable.
My problem is in the second part. If $lambda: mathbb{R} to mathbb{R}^{2}$ is a line pararallel to $e_{i}$ passing through $0$,
$lambda(t) = (0,0) + te_{i}$, I can show that $(fcirc lambda)'(0) = frac{partial}{partial x_{i}}f(0,0)$. I've tried to generalize this to an arbitrary path with domain $(0,1)$, but I couldnt. Can someone help me?
real-analysis derivatives
real-analysis derivatives
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
edited Jan 19 at 16:16
Lucas Corrêa
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
asked Jan 19 at 15:59
Lucas CorrêaLucas Corrêa
1,5871321
1,5871321
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16
add a comment |
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Denote $lambda(t) = (lambda_1(t), lambda_2(t))$. Since $lambda$ goes through the origin, we have that $lambda(c) = (0,0)$ for some $cin(0,1)$. We can assume that $lambda'(c) ne 0$ so that $lambda$ isn't constant on a neighbourhood of the origin.
Then $(fcirc lambda)(c) = 0$ so by definition:
$$(fcirc lambda)'(c) = lim_{tto 0} frac{(fcirc lambda)(c+t)- (fcirc lambda)(c)}t = lim_{t to 0} frac{(fcirc lambda)(c+t)}t= lim_{t to 0} frac{lambda_1(c+t)^3}{t(lambda_1(c+t)^2 + lambda_2(c+t)^2)}$$
Notice that
$$lim_{tto 0} frac{lambda_1(c+t)}{t} = lim_{tto 0} frac{lambda_1(c+t)-lambda_1(c)}{t}= lambda_1'(c)$$
and similarly for $lambda_2$.
Therefore
$$(fcirc lambda)'(c) = lim_{tto 0} frac{lambda_1(c+t)^3}{t^3} cdot frac{t^2}{lambda_1(c+t)^2 + lambda_2(c+t)^2}=lim_{tto 0} left(frac{lambda_1(c+t)}{t}right)^3 cdot frac1{left(frac{lambda_1(c+t)}{t}right)^2 + left(frac{lambda_2(c+t)}{t}right)^2}$$
Since $lambda'(c) ne 0$, the product rule for limits yields $$(fcirc lambda)'(c) = frac{lambda_1'(c)^3}{lambda_1'(c)^2 + lambda_2'(c)^2}$$
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
$endgroup$
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
|
show 4 more comments
$begingroup$
A broad hint, but (deliberately) not a complete solution
Suppose that $alpha$ and $beta$ are two differentiable paths with $alpha(0) = beta(0) = (0,0)$ and $alpha'(0) = beta'(0) ne 0$. In short form, '$alpha$ and $beta$ agree to first order at $0$'.
You can prove that if $f circ alpha$ is differentiable at $0$, then so is $f circ beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $alpha'(0)$ is nonzero will be essential.)
Once you know that, you need only consider a few possible curves through the origin, namely those of the form
$$
alpha(t) = t mathbf{v}
$$
where $mathbf{v} = (a, b) = r (cos theta, sin theta)
$
is some nonzero vector, i.e., $r ne 0$. For each such vector, it's easy to write down $f circ alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3079497%2fdifferentiability-through-paths%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Denote $lambda(t) = (lambda_1(t), lambda_2(t))$. Since $lambda$ goes through the origin, we have that $lambda(c) = (0,0)$ for some $cin(0,1)$. We can assume that $lambda'(c) ne 0$ so that $lambda$ isn't constant on a neighbourhood of the origin.
Then $(fcirc lambda)(c) = 0$ so by definition:
$$(fcirc lambda)'(c) = lim_{tto 0} frac{(fcirc lambda)(c+t)- (fcirc lambda)(c)}t = lim_{t to 0} frac{(fcirc lambda)(c+t)}t= lim_{t to 0} frac{lambda_1(c+t)^3}{t(lambda_1(c+t)^2 + lambda_2(c+t)^2)}$$
Notice that
$$lim_{tto 0} frac{lambda_1(c+t)}{t} = lim_{tto 0} frac{lambda_1(c+t)-lambda_1(c)}{t}= lambda_1'(c)$$
and similarly for $lambda_2$.
Therefore
$$(fcirc lambda)'(c) = lim_{tto 0} frac{lambda_1(c+t)^3}{t^3} cdot frac{t^2}{lambda_1(c+t)^2 + lambda_2(c+t)^2}=lim_{tto 0} left(frac{lambda_1(c+t)}{t}right)^3 cdot frac1{left(frac{lambda_1(c+t)}{t}right)^2 + left(frac{lambda_2(c+t)}{t}right)^2}$$
Since $lambda'(c) ne 0$, the product rule for limits yields $$(fcirc lambda)'(c) = frac{lambda_1'(c)^3}{lambda_1'(c)^2 + lambda_2'(c)^2}$$
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
$endgroup$
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
|
show 4 more comments
$begingroup$
Denote $lambda(t) = (lambda_1(t), lambda_2(t))$. Since $lambda$ goes through the origin, we have that $lambda(c) = (0,0)$ for some $cin(0,1)$. We can assume that $lambda'(c) ne 0$ so that $lambda$ isn't constant on a neighbourhood of the origin.
Then $(fcirc lambda)(c) = 0$ so by definition:
$$(fcirc lambda)'(c) = lim_{tto 0} frac{(fcirc lambda)(c+t)- (fcirc lambda)(c)}t = lim_{t to 0} frac{(fcirc lambda)(c+t)}t= lim_{t to 0} frac{lambda_1(c+t)^3}{t(lambda_1(c+t)^2 + lambda_2(c+t)^2)}$$
Notice that
$$lim_{tto 0} frac{lambda_1(c+t)}{t} = lim_{tto 0} frac{lambda_1(c+t)-lambda_1(c)}{t}= lambda_1'(c)$$
and similarly for $lambda_2$.
Therefore
$$(fcirc lambda)'(c) = lim_{tto 0} frac{lambda_1(c+t)^3}{t^3} cdot frac{t^2}{lambda_1(c+t)^2 + lambda_2(c+t)^2}=lim_{tto 0} left(frac{lambda_1(c+t)}{t}right)^3 cdot frac1{left(frac{lambda_1(c+t)}{t}right)^2 + left(frac{lambda_2(c+t)}{t}right)^2}$$
Since $lambda'(c) ne 0$, the product rule for limits yields $$(fcirc lambda)'(c) = frac{lambda_1'(c)^3}{lambda_1'(c)^2 + lambda_2'(c)^2}$$
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
$endgroup$
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
|
show 4 more comments
$begingroup$
Denote $lambda(t) = (lambda_1(t), lambda_2(t))$. Since $lambda$ goes through the origin, we have that $lambda(c) = (0,0)$ for some $cin(0,1)$. We can assume that $lambda'(c) ne 0$ so that $lambda$ isn't constant on a neighbourhood of the origin.
Then $(fcirc lambda)(c) = 0$ so by definition:
$$(fcirc lambda)'(c) = lim_{tto 0} frac{(fcirc lambda)(c+t)- (fcirc lambda)(c)}t = lim_{t to 0} frac{(fcirc lambda)(c+t)}t= lim_{t to 0} frac{lambda_1(c+t)^3}{t(lambda_1(c+t)^2 + lambda_2(c+t)^2)}$$
Notice that
$$lim_{tto 0} frac{lambda_1(c+t)}{t} = lim_{tto 0} frac{lambda_1(c+t)-lambda_1(c)}{t}= lambda_1'(c)$$
and similarly for $lambda_2$.
Therefore
$$(fcirc lambda)'(c) = lim_{tto 0} frac{lambda_1(c+t)^3}{t^3} cdot frac{t^2}{lambda_1(c+t)^2 + lambda_2(c+t)^2}=lim_{tto 0} left(frac{lambda_1(c+t)}{t}right)^3 cdot frac1{left(frac{lambda_1(c+t)}{t}right)^2 + left(frac{lambda_2(c+t)}{t}right)^2}$$
Since $lambda'(c) ne 0$, the product rule for limits yields $$(fcirc lambda)'(c) = frac{lambda_1'(c)^3}{lambda_1'(c)^2 + lambda_2'(c)^2}$$
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
$endgroup$
Denote $lambda(t) = (lambda_1(t), lambda_2(t))$. Since $lambda$ goes through the origin, we have that $lambda(c) = (0,0)$ for some $cin(0,1)$. We can assume that $lambda'(c) ne 0$ so that $lambda$ isn't constant on a neighbourhood of the origin.
Then $(fcirc lambda)(c) = 0$ so by definition:
$$(fcirc lambda)'(c) = lim_{tto 0} frac{(fcirc lambda)(c+t)- (fcirc lambda)(c)}t = lim_{t to 0} frac{(fcirc lambda)(c+t)}t= lim_{t to 0} frac{lambda_1(c+t)^3}{t(lambda_1(c+t)^2 + lambda_2(c+t)^2)}$$
Notice that
$$lim_{tto 0} frac{lambda_1(c+t)}{t} = lim_{tto 0} frac{lambda_1(c+t)-lambda_1(c)}{t}= lambda_1'(c)$$
and similarly for $lambda_2$.
Therefore
$$(fcirc lambda)'(c) = lim_{tto 0} frac{lambda_1(c+t)^3}{t^3} cdot frac{t^2}{lambda_1(c+t)^2 + lambda_2(c+t)^2}=lim_{tto 0} left(frac{lambda_1(c+t)}{t}right)^3 cdot frac1{left(frac{lambda_1(c+t)}{t}right)^2 + left(frac{lambda_2(c+t)}{t}right)^2}$$
Since $lambda'(c) ne 0$, the product rule for limits yields $$(fcirc lambda)'(c) = frac{lambda_1'(c)^3}{lambda_1'(c)^2 + lambda_2'(c)^2}$$
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
edited Jan 20 at 19:02
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
answered Jan 19 at 16:33
mechanodroidmechanodroid
27.9k62447
27.9k62447
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
|
show 4 more comments
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
I would say, near the bottom, "Because $lambda'(c) ne 0$, the product rule for limits yields"
$endgroup$
– John Hughes
Jan 19 at 18:53
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
@JohnHughes Note that I already assumed that $lambda'(c) ne 0$ at the beginning
$endgroup$
– mechanodroid
Jan 19 at 18:57
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
I did note that. I was making a suggestion about how one might present the material clearly to the OP, not on completeness of the argument. Right now, you make that assumption, but never mention it again in the proof, and the reader might wonder, "Why was that assumption there at all?" This is clearly a matter of taste. For the right reader, the proof might be reduced to "Any fool can plainly see that this is true" and be entirely adequate.
$endgroup$
– John Hughes
Jan 20 at 16:13
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
@JohnHughes Ok, you may be right, I added it in the answer. However, I'm not too happy with my argument why $lambda'(c) ne 0$ in the first place. Is there a clearer explanation why we should only consider paths with nonzero derivative at the origin?
$endgroup$
– mechanodroid
Jan 20 at 19:03
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
$begingroup$
Two arguments, I think: One is that the lemma I wrote (if $alpha, beta$ agree to first order at $t_0$, then $fcirc alpha$ and $f circ beta$ have the same derivative at $t_0$, only holds when $alpha'(t_0) ne 0$ in general. But for this particular function $f$, if $lambda(t_0) = (0,0)$ and $lambda'(t_0) = 0$, then $(fcirc lambda)'(t_0) = 0$ (which you can prove by computing limits pretty easily, I believe). So really the answer is: the zero-speed case is another case you should handle, but it's not too tough.
$endgroup$
– John Hughes
Jan 20 at 19:39
|
show 4 more comments
$begingroup$
A broad hint, but (deliberately) not a complete solution
Suppose that $alpha$ and $beta$ are two differentiable paths with $alpha(0) = beta(0) = (0,0)$ and $alpha'(0) = beta'(0) ne 0$. In short form, '$alpha$ and $beta$ agree to first order at $0$'.
You can prove that if $f circ alpha$ is differentiable at $0$, then so is $f circ beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $alpha'(0)$ is nonzero will be essential.)
Once you know that, you need only consider a few possible curves through the origin, namely those of the form
$$
alpha(t) = t mathbf{v}
$$
where $mathbf{v} = (a, b) = r (cos theta, sin theta)
$
is some nonzero vector, i.e., $r ne 0$. For each such vector, it's easy to write down $f circ alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
add a comment |
$begingroup$
A broad hint, but (deliberately) not a complete solution
Suppose that $alpha$ and $beta$ are two differentiable paths with $alpha(0) = beta(0) = (0,0)$ and $alpha'(0) = beta'(0) ne 0$. In short form, '$alpha$ and $beta$ agree to first order at $0$'.
You can prove that if $f circ alpha$ is differentiable at $0$, then so is $f circ beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $alpha'(0)$ is nonzero will be essential.)
Once you know that, you need only consider a few possible curves through the origin, namely those of the form
$$
alpha(t) = t mathbf{v}
$$
where $mathbf{v} = (a, b) = r (cos theta, sin theta)
$
is some nonzero vector, i.e., $r ne 0$. For each such vector, it's easy to write down $f circ alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
$endgroup$
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
add a comment |
$begingroup$
A broad hint, but (deliberately) not a complete solution
Suppose that $alpha$ and $beta$ are two differentiable paths with $alpha(0) = beta(0) = (0,0)$ and $alpha'(0) = beta'(0) ne 0$. In short form, '$alpha$ and $beta$ agree to first order at $0$'.
You can prove that if $f circ alpha$ is differentiable at $0$, then so is $f circ beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $alpha'(0)$ is nonzero will be essential.)
Once you know that, you need only consider a few possible curves through the origin, namely those of the form
$$
alpha(t) = t mathbf{v}
$$
where $mathbf{v} = (a, b) = r (cos theta, sin theta)
$
is some nonzero vector, i.e., $r ne 0$. For each such vector, it's easy to write down $f circ alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
$endgroup$
A broad hint, but (deliberately) not a complete solution
Suppose that $alpha$ and $beta$ are two differentiable paths with $alpha(0) = beta(0) = (0,0)$ and $alpha'(0) = beta'(0) ne 0$. In short form, '$alpha$ and $beta$ agree to first order at $0$'.
You can prove that if $f circ alpha$ is differentiable at $0$, then so is $f circ beta$, and the derivatives agree. (I think I'd use an epsilon-delta argument to show this. The proof should not depend on $f$, by the way, but the assumption that the derivative $alpha'(0)$ is nonzero will be essential.)
Once you know that, you need only consider a few possible curves through the origin, namely those of the form
$$
alpha(t) = t mathbf{v}
$$
where $mathbf{v} = (a, b) = r (cos theta, sin theta)
$
is some nonzero vector, i.e., $r ne 0$. For each such vector, it's easy to write down $f circ alpha(t)$ and observe that it's differentiable, and then (with the help of the second paragraph) you're done.
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
answered Jan 19 at 16:18
John HughesJohn Hughes
64.4k24191
64.4k24191
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
add a comment |
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
Its a nice hint, John! I'll try to use it. Just a point: the domain of path not contains $0$ so, is it still possible to keep the same approach with some trick?
$endgroup$
– Lucas Corrêa
Jan 19 at 16:26
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
The domain of my paths $alpha$ and $beta$ are both supposed to be some open interval containing $0$, such as $-1 < t < 1$ (or perhaps all of $Bbb R$). For the particular path $alpha$ that I chose ($alpha(t) = tmathbf{v}$), the domain might as well be all of $Bbb R$, just as it was for your path $lambda$. In short, I don't know what you're asking here, because we seem to be doing the same thing in general, except that instead of just considering the vertical and horizontal lines through the origin, I'm considering all lines through the origin.
$endgroup$
– John Hughes
Jan 19 at 16:31
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
$begingroup$
Yes, I understood your idea. In fact, as I said, I was able to solve a very particular case considering $mathbb{R}$. But the question asks for a path with domain $(0,1)$. Thus, at least my approach, doesnt work anymore.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:35
1
1
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
$begingroup$
Just replace my definition of $alpha$ with $alpha(t) = (t - frac12) mathbf{v}$, and do all the computations at $t = frac12$. (Away from $t = frac12$, $f$ is everywhere differentiable so the chain rule shows things are nice. At $t = frac12$, you have to use limits, etc.)
$endgroup$
– John Hughes
Jan 19 at 16:58
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3079497%2fdifferentiability-through-paths%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What does path-differentiable mean?There is the usual definition of the derivative as a linear transformation satisfying some condition, and differentiability meaning that the derivative exists.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:14
$begingroup$
It got badly written, I'll change it. It is only a path that is differentiable.
$endgroup$
– Lucas Corrêa
Jan 19 at 16:15
$begingroup$
Okay, thank you.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 16:16