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While attempting to answer this question with help of MAPLE, something very strange happened, at least according to me. We have the following function: $$ \phi(x,y) = x^2-y^2+\frac{x^2-y^2}{(x^2+y^2)^2} $$ The partial derivatives of this function are (according to MAPLE and myself): $$ u = \frac{\partial \phi}{\partial x} = 2 x + \frac{2 x}{(x^2+y^2)^2} - \frac{4(x^2-y^2)x}{(x^2+y^2)^3} \\ v = \frac{\partial \phi}{\partial y} = - 2 y - \frac{2 y}{(x^2+y^2)^2} - \frac{4(x^2-y^2)y}{(x^2+y^2)^3} $$ Then we calculate $\sqrt{u^2+v^2}$ and ask MAPLE to simplify, with the following result: $$ \sqrt{u^2+v^2} = 2\,\sqrt {{\frac{x^8+4\,x^6y^2-2\,x^4+6\,x^4y^4 +12\,x^2y^2+4\,x^2y^6-2\,y^4+y^8 \color{red}{+1}}{\left(x^2+y^2\right)^{3}}}} $$ But, even at first sight .. how can there be a $+1$ in that fraction ? Here comes the MAPLE sequence: 

phi(x,y) := (x^2-y^2)+(x^2-y^2)/(x^2+y^2)^2;
diff(phi(x,y),x); diff(phi(x,y),y);
simplify(sqrt(diff(phi(x,y),x)^2+diff(phi(x,y),y)^2));
I use a legal MAPLE 8 version, to be precise.
DSkoog
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Han de Bruijn
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    Why wouldn't there be a constant term in the numerator? Let $y=0$. Then $v=0$ and $u^2=4(x^4-1)^2/x^6$ so there must be a constant term. – WimC Jun 08 '14 at 19:31

3 Answers3

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Nothing wrong here, for example if you set $y=0$ and simplify by hand, you get

$$\sqrt{u^2} = 2\sqrt{\frac{(x^4-1)^2}{x^6}}$$

exactly what Maple says. You can see how $1$ emerges...

PA6OTA
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Maple 18 gives the same answer, which, as far as I can tell, is correct. For example, at $x=3, y=7$ both $\sqrt{u^2+v^2}$ calculated directly and Maple's answer simplify to $15\sqrt{50297} \sqrt{58}/1682$.

Robert Israel
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Here is a link for Maple bugs:

http://www.math.ncsu.edu/MapleInfo/MapleArchive/BugList.html

This is for maple V, but it was only a quick search. The moral of the story is be careful, and don't be so quick to discount drawnonward. I have had problems with Maple before....years ago. And I have a great memory. Oh, and I think all those down votes we certainly a bit hasty, as was my original answer. I apologize.

drawnonward
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  • No hard feelings and (+1) for this answer. OK? – Han de Bruijn Jun 08 '14 at 19:56
  • Non at all...ty. The fact of the matter is that algorithms are used to do these calculations, so anyone that thinks they are perfect is way off. Calculators use mac series to calculate trig. functions, and at the risk of getting down voted again, I have had calculation errors there as well....technology isn't perfect, and you weren't wrong to ask if it was an error. – drawnonward Jun 08 '14 at 19:57
  • As for the others that down voted me, they are forgiven too. I was being argumentative. Although @user61527 clearly you are the one who is "pretty much wrong" – drawnonward Jun 08 '14 at 20:03