The approximation function of $\frac{x}{y}$

Question

Is there a approximation function of $$\frac{x}{y},$$ and the approximation function is in the form of $f(x) + f(y)$ or $f(x) - f(y)$. That's to say the approximation function can split $x$ and $y$.

Not sure what you want. Since $\frac xx=1$ (at least for $x\neq 0$) we need $f(x)+f(x)=1$ or $f(x)-f(x)=1$, neither of which seems likely to lead anywhere. — lulu, May 30 '17 at 11:16
What are the possible values for $x$ and $y$ ? There will probably not be an approximation of the desired type good for all pairs $(x,y)$ — Peter, May 30 '17 at 11:20
There is no such approximation in any neighborhood of a point at which $y=0$ and $x\ne 0,$ since the quotient approaches $\infty$ at such points. At other points where $x\ne 0,$ there is a tangent plane, and that gives you a linear approximation. $\qquad$ — Michael Hardy, May 30 '17 at 11:34
@MichaelHardy Thanks a lot, and what's the linear approximation when $x \ne 0$. — UniMilky, May 30 '17 at 12:29
@UniMilky : Let's try the point $(x,y)=(1,1).$ then $\dfrac\partial{\partial x} ,\dfrac x y = \dfrac 1 y$ and $\dfrac\partial{\partial y} , \dfrac x y = \dfrac{-x}{y^2}.$ The values of those two derivatives at $(x,y)=(1,1)$ are respectively $1$ and $-1$. Therefore $\Delta\dfrac x y \approx 1\cdot\Delta x + (-1)\cdot\Delta y.$ Since the value of $x/y$ at $(x,y)=(1,1)$ is $1$, that yields $$ \begin{align} \frac x y & \approx 1 + 1\cdot\Delta x + (-1)\cdot\Delta y \ \ & = 1 + 1(x-1) + (-1)(y-1). \end{align} $$ — Michael Hardy, May 30 '17 at 12:35

score 0 · Answer 1 · answered May 30 '17 at 11:48

Though the question is unspecific about what constitutes an "approximation", the answer appears to be "no".

As lulu notes in the comments, an approximation $\frac{x}{y} \approx f(x) + f(y)$ leads (for $x = y$) to $$ 1 = \frac{x}{x} \approx f(x) + f(x) = 2f(x)\quad\text{for all $x$.} $$

Similarly, an approximation $\frac{x}{y} \approx f(x) - f(y)$ leads (for $x = y$) to $$ 1 = \frac{x}{x} \approx f(x) - f(x) = 0. $$

From the other direction (i.e., starting with customary notions of approximation and seeing where they lead):

If $y_{0} \neq 0$, then for $|y - y_{0}| < |y_{0}|$ the geometric series gives the first-order approximation \begin{align*} \frac{x}{y} &= \frac{x}{y_{0} + (y - y_{0})} = \frac{x}{y_{0}} \cdot \frac{1}{1 + (\frac{y - y_{0}}{y_{0}})} \\ &= \frac{x}{y_{0}} \cdot \left[1 - \frac{y - y_{0}}{y_{0}} + \bigg(\frac{y - y_{0}}{y_{0}}\biggr)^{2} - \cdots\right] \\ &\approx \frac{x}{y_{0}} - \frac{x(y - y_{0})}{y_{0}^{2}}, \end{align*} which is not of the form you seek.

The idea is clearly,Thanks a lot. – UniMilky May 30 '17 at 12:22 — UniMilky, May 30 '17 at 12:22

The approximation function of $\frac{x}{y}$

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