I tried to do that using this method :

let $f(x,y) = x^y$ so now I'm looking for $f(0.93,2.98)$

$\Delta f = f(1,3)-f(0.93,2.98) \approx df = f_xdx+f_ydy$

here $x=1,y=3,dx=-0.07,dy=-0.02$

$f_x=yx^{y-1},f_y=x^y\ln x,f(1,3)=1$

so $$df = 1\cdot(-0.07)+0\cdot(-0.02) = -0.07$$

$$df = -0.07\approx f(1,3)-f(0.93,2.98) \implies 0.93^{2.98} \approx 1.07 $$

but the result I got is a bit far from the real value of $0.93^{2.98}$ ($0.80533$)

my question is : is my reasoning false or is it just a bad approximation ?

  • $\begingroup$ You have $f_x = 1$, but you should have $f_x=3$. This will give you a better approximation of $0.79$ $\endgroup$ – AlexanderJ93 Apr 13 '17 at 8:40
  • $\begingroup$ Check value of $f_x$ and also you have sign mistake. $\endgroup$ – user121049 Apr 13 '17 at 8:41
  • $\begingroup$ @ AlexanderJ93 my bad I plugged the values into $f(x,y)$ not $f_x$ thanks for noticing @user121049 thanks for answering but I don't see where I did a sign mistake can you please point it out ? $\endgroup$ – the_firehawk Apr 13 '17 at 8:44
  • $\begingroup$ @rapidracim I don´t see a sign mistake as well. $\endgroup$ – callculus Apr 13 '17 at 8:49
  • 1
    $\begingroup$ Well, the "error in sign" is in writing $$f(1,3)-f(0.98,2.93)\approx (0.93-1)\frac{\partial f}{\partial x}(1,3)+(2.98-3)\frac{\partial f}{\partial y}(1,3)$$ Taylor expansion would be either $$f(0.98,2.93)-f(1,3)\approx (0.93-1)\frac{\partial f}{\partial x}(1,3)+(2.98-3)\frac{\partial f}{\partial y}(1,3)$$ or $$f(1,3)-f(0.93,2.98)\approx (1-0.93)\frac{\partial f}{\partial x}(0.93,2.98)+(3-2.98)\frac{\partial f}{\partial y}(0.93,2.98)$$ (of course, the last one is not the one we're looking for) $\endgroup$ – user228113 Apr 13 '17 at 9:12

The method is fine.

$f_x=yx^{y-1},f_y=x^y\ln x,f(1,3)=1$

You have $f_x(x,y)=yx^{y-1} \implies f_x(1,3)=3 \cdot 1^{2} = 3$; so

$$df = \color{red}{1}\cdot(-0.07)+0\cdot(-0.02) = -0.07$$

should be:

$$df = \color{green}{3}\cdot(-0.07)+0\cdot(-0.02) = \color{blue}{-0.21}$$

which gives:

$$0.93^{2.98} = f(0.93,2.98) \approx f(1,3)\color{blue}{-0.21} = 0.79$$

  • $\begingroup$ A bit late I see; typesetting and coloring takes longer than commenting ;-). $\endgroup$ – StackTD Apr 13 '17 at 8:48

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