Economics Job Market Rumors Topic: Serious calculus question
https://www.econjobrumors.com/topic/serious-calculus-question
Economics Job Market Rumors Topic: Serious calculus questionen-USWed, 07 Jun 2023 18:48:02 +0000http://bbpress.org/?v=1.0.2<![CDATA[Search]]>q
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Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-429717
Thu, 14 Jun 2012 17:26:06 +0000Economist429717@https://www.econjobrumors.com/<blockquote><p>OP's statement is correct. If the function is differentiable (at (x,y)), then one can replace a directional derivative (LHS) with the product of the gradient and the direction vector (RHS). It is true that the direction vector is often taken to have unit length, while here it is (1,1), so that the LHS is a directional derivative scaled by 2^0.5 but the scaling occurs on the RHS as well, and thus the equality still holds.</p></blockquote>
<p>This is 100% correct.
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-429626
Thu, 14 Jun 2012 15:12:40 +0000Economist429626@https://www.econjobrumors.com/<p>ditto, OP is correct whenever the function is differentiable (in two dimensions) at (x,y). A sufficient condition is that both partial derivatives are continuous at the point. Obviously it cannot be true in general.
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-429561
Thu, 14 Jun 2012 13:52:05 +0000Economist429561@https://www.econjobrumors.com/<blockquote><p>Perhaps a silly question, but suppose I have a differentiable function f(x,y), is the following statement true?<br />
\lim_{\varepsilon \rightarrow 0} \frac{f(x+\varepsilon,y+\varepsilon)}{\varepsilon}=\lim_{\varepsilon \rightarrow 0} \frac{f(x+\varepsilon,y)}{\varepsilon}+\lim_{\varepsilon \rightarrow 0} \frac{f(x,y+\varepsilon)}{\varepsilon}</p></blockquote>
<p>Ok assuming f(x,y)=0 then you are asking if a directional derivative (45 degree angle) is equal to the sum of the partial derivatives. Why would that always be true? It makes no sense at all.
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-429413
Thu, 14 Jun 2012 08:01:33 +0000Economist429413@https://www.econjobrumors.com/<p>Look up total differentiability.</p>
<p>You'll get into trouble with functions like f(x,y)=min(x,2y) at (0,0).
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-429057
Wed, 13 Jun 2012 21:19:31 +0000Economist429057@https://www.econjobrumors.com/<p>TT is laughing
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question/page/2#post-428995
Wed, 13 Jun 2012 19:55:33 +0000Economist428995@https://www.econjobrumors.com/<p>In my browser even the question has vanished, I call that really taking things to the limit.
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question#post-428989
Wed, 13 Jun 2012 19:44:42 +0000Economist428989@https://www.econjobrumors.com/<p>I think you need more than differentiability (and f(x,y)=0). Do you need it to be continuously differentiable at (x,y)?
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question#post-428945
Wed, 13 Jun 2012 18:42:55 +0000Economist428945@https://www.econjobrumors.com/<blockquote><p>OP's statement is correct. If the function is differentiable (at (x,y)), then one can replace a directional derivative (LHS) with the product of the gradient and the direction vector (RHS). It is true that the direction vector is often taken to have unit length, while here it is (1,1), so that the LHS is a directional derivative scaled by 2^0.5 but the scaling occurs on the RHS as well, and thus the equality still holds.</p></blockquote>
<p>Thank you! Finally a competent answer.
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question#post-428940
Wed, 13 Jun 2012 18:40:05 +0000Economist428940@https://www.econjobrumors.com/<blockquote><p>OK moron. You are obviously a troll. Carry out the first limit under the assumption that f is x+y. You got a two?</p></blockquote>
<p>Yeah, as</p>
<p><bblatex>\lim_{\varepsilon \rightarrow 0} \frac{x+\varepsilon+y+\varepsilon-x-y}{\varepsilon}=2</bblatex></p>
<p>Any other insights?
</p>Economist on "Serious calculus question"
https://www.econjobrumors.com/topic/serious-calculus-question#post-428939
Wed, 13 Jun 2012 18:38:42 +0000Economist428939@https://www.econjobrumors.com/<p>... and, of course, assuming that f(x,y)=0, as has been pointed out.
</p>