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I found the quadratic approximation as $9 + \frac{1}{2}(-9x^2 - 9y^2)$

The problem is that the triple derivatives all end up 0 at (0,0), so I get that the error approximation is 0. Wolfram alpha calculates the triple derivatives having sin(y) or sin(x) in them making them 0. I know you are supposed to plug in the values into $1/6(f_(xxx)(x^3) + ...)$ but each triple derivative ends up as 0 which turns out to be the wrong answer.

Ross Millikan
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    The qudratic approximation happens to be the same as the cubic approximation. So for the error analysis you can use the cubic approximation. Either do that, or use the third partials, and pessimistic estimates for the sines in the third partials. – André Nicolas Apr 03 '16 at 00:16
  • The problem is that the cubic approximation makes the error zero though. –  Apr 03 '16 at 01:26
  • The problem is sin(0) is 0. I heard that the max is taken over an interval, so I shouldn't be using 0. What number should I plug in to the third dervitaves when getting the error analysis? –  Apr 03 '16 at 01:29
  • The things that make the functions worst, so for $\sin(x)$ use $\sin(0.27)$, for $\cos x$ use $1$. – André Nicolas Apr 03 '16 at 01:35
  • I see. In this case, it should be 9 for the third partials because 9 is a multiple. I got the right answer that way. –  Apr 03 '16 at 01:42

2 Answers2

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$f(x+h,y+k) \approx f(x,y) + h \dfrac{\partial f}{\partial x} = k \dfrac{\partial f}{\partial y} + \dfrac 12 \left[ h^2 \dfrac{\partial^2 f}{\partial x^2} + 2hk \dfrac{\partial^2 f}{\partial x \partial y} + k^2 \dfrac{\partial^2 f}{\partial y^2} \right]=$

$9 \cos x \cos y + -h \cdot 9 \sin x \cos y -k \cdot 9 \cos x \sin y$
$+ \dfrac 12 \left[ -h^2 \cdot \cos x \cos y + 2hk \cdot \sin x \sin y + -k^2 \cdot \cos x \cos y \right]$

At $x = 0,\; y=0$, this becomes $f(0+h,0+k) \approx 9 - \dfrac 12[h^2 + k^2]$.

For $h = 0.27$, and $k = 0.03$ we get $|f(h,k) - 9| \approx \dfrac 12[0.27^2 + 0.03^2]$.
$|f(h,k) - f(0,0)| \approx 0.0369$

Steven Alexis Gregory
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Since the cosine series is alternating one has $$1-{x^2\over2}<\cos x<1-{x^2\over2}+{x^4\over24}$$ when $|x|$ is small. It follows that $$0.96355=1-{1\over2}0.27^2<\cos0.27<0.96355+0.00022\ ,$$ and in the same way $$0.99955=1-{1\over2}0.03^2<\cos0.03=0.99955+4\cdot10^{-8}\ .$$ From this we can conclude that $$0.9631164<\cos0.27\cdot\cos0.03<0.963338\ ,$$ or $$8.668047<9\cos0.27\cdot\cos0.03<8.67004\ .$$ Taking the mean between the two bounds gives $8.669044$ with an error $<0.001$. Mathematica produces $8.67003508790$.

Christian Blatter
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