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Why does one need to use $f(1,0)$ a linear approximation, rather than calculating $f(1.1, -0.1)$ directly?

e.g. when $f(x,y)=xe^{xy}$ (used to be more complicated $e^{xy}+xye^{xy}$)

It's easy to see that $f$ is a not a linear function. however I cannot recall, why is it uncomputable on fractional points?

Example given here:

https://sites.ualberta.ca/~rjia/Math214/Part4/Lec6.pdf

mavavilj
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    You don't need to, since in this case you can easily compute the exact value. Such simple examples are just for teaching purposes, to illustrate the idea of linear approximation (which is extremely useful in other contexts). – Hans Lundmark Oct 11 '17 at 15:00
  • @HansLundmark I was thinking that might be the case. – mavavilj Oct 11 '17 at 15:00
  • May you have a look to the explanation belows – Guy Fsone Oct 11 '17 at 15:24
  • In a "real-life" problem, the value (or the equation) of $f(x,y)$ may not be known everywhere but the values (or the equations) of the partial derivative are known. – Steven Alexis Gregory Oct 11 '17 at 15:44

2 Answers2

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Note: I did this using your original function: $f(x,y)=e^{xy}+xye^{xy}$, the point still stands.

This result is quite computable, but it may be impractical to compute. It may be more worth while to simply get a quite very strong approximation. I am by no means experienced with these sorts of things, but when it comes to some types of physics or engineering, sometimes a linear approximation is good enough for the purpose. Worrying about the specific value can be just a waste of time and computing power in practice.

In this case, we have by approximation that $$f(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0). $$ So, if we approximate at $(1,0)$, we have $$ e^{xy}+xye^{xy}\approx 1+(2ye^{xy}+xy^2e^{xy})\bigg|_{(1,0)}(x-1)+ (2xe^{xy}+x^2ye^{xy})\bigg|_{(1,0)}(y-0)=1+2y.$$ So, we have that $f(1.1,-0.1)\approx 1+2(-0.1)=.8$. If we use Wolfram Alpha to compute the result directly, we find that $f(1.1,-0.1)=.79729...$ so that our error is extremely small.

Alekos Robotis
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It is fact the matter of choice.

  • The value $f(1,0) = 1$ is known
  • $(1, 0)$ is the closest element to (1.1,-0.1) amount $(x,y)$ such that $ f(x,y)$ is that is explicitly known. $$|(1.1,-01)-(1,0)| =|(0.1,-0.1)|1= \sqrt2/10$$
  • $$f(1.1,-0.1) = f(1+0.1, 0-0.1) \sim f(1,0) +\nabla f(1,0)\cdot (0.1,-0.1) = 1+(1,1)\cdot (0.1,-0.1) = 1 $$

But one could chose any other value $(x_0,y_0)$ even if it far from $(1.1,-0.1) $ But be aware that in this case you will need to take the Taylor expansion more further in other to get the right approximation. for instance you could chose $$(2,0)\implies f(2,0) =2$$ but to get the suitable approximation of $$f(1.1,-0.1)$$ with the expansion around $$(2,0)$$ you will need more oder in your expansion than when you use $$(1,0)$$ Because $(1,0)$ is more closer to $(1.1, -0.1)$ than $(2,0)$. But also that you could actually use $$(-1,0)\implies f( -1,0) =-1$$

$(-1,0) $ is still close to $(1.1,-0.1)$ as well BUT we have

$$f(1.1,-0.1) = f(-1+2.1, 0-0.1) \sim f(-1,0) +\nabla f(-1,0)\cdot (2.1,-0.1) \\= -1+(1,1)\cdot (2.1,-0.1) = -1 + 2.1-0.1 = 1 $$

So with $(-1,0)$ and $(1,0)$ one obatain the same aprroximation with the first oder Taylor expansion

Guy Fsone
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