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

Question

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

Answer 1

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.

Answer 2

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