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Suppose we have a function $f$ of $bx-ay$ where $a$ and $b$ are two real constants, if we have for example $e^{bx-ay}$ then obviously it is a function of $bx-ay$.

Can we find a function $f$ such that: $f(bx-ay) = ax-by$? in other words what operations we should operate on $bx-ay$ to get $ax-by$? how can I proceed?

Syntax_ErrorX00
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2 Answers2

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No. Here's why. Consider $a = 1, b = -2$ and look at $x = 0, y = 2$ and $x = 1, y = 0$. For each of these $bx - ay = -2$. But for the first, $ax - by = 4$, while for the second $ax - by = 1$.

Since the function $f$ can only take on one value for the argument $-2$ (because of the definition of "function"), it must take on the value either $4$ or $1$, but not both.

A similar argument works for almost any other pair of values for $a$ and $b$; the only exception I can see is $a = b = 0$, when it's easy to build $f$ (but not interesting: it's the everywhere-zero function).

John Hughes
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  • It is possible when $a=b$ or when $a=-b$. – hardmath Jun 29 '16 at 02:26
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    In fact you can get any $a=b$. For example $a=b=2$, $f(2x-2y)=2x-2y$ is equivalent to $f(x)=-x$ – Andrei Jun 29 '16 at 02:27
  • @JohnHughes: is there any other methods, say, chain rule or something similar to use in general case? – Syntax_ErrorX00 Jun 29 '16 at 02:33
  • The chain rule might be helpful...if you knew that the function $f$ were differentiable. But that's assuming a lot. Thanks, Andrei and Hardmath, for pointing out the other cases where there are trivial solutions -- I should have seen those! – John Hughes Jun 29 '16 at 02:38
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Here is my solution:

we can put $z = bx-ay $ and $w = ax-by$ So we have: $$f(z) = w$$ from the previous relation we get: \begin{cases} x = \dfrac{aw-bz}{a^2-b^2}\\ y = \dfrac{bw-az}{a^2-b^2} \end{cases}

if $w$ do not depends on $z$ then $\dfrac{\partial w}{\partial z}=0$

hence: $$\dfrac{\partial w }{\partial z} = \dfrac{\partial w }{\partial x }\dfrac{\partial x}{\partial z }+\dfrac{\partial w}{\partial y}\dfrac{\partial y }{\partial z } = 0$$

we have: $\dfrac{\partial w}{\partial x } = a$, $\dfrac{\partial w}{\partial y } = -b$, $\dfrac{\partial x}{\partial z } = \dfrac{-b }{a^2-b^2 }$ and $\dfrac{\partial y }{\partial z } = \dfrac{-a }{a^2-b^2 }$

Therefore: $$\dfrac{-ab }{a^2-b^2 }+\dfrac{ab }{a^2-b^2 }=0$$ This is valid only if $a^2\neq b^2 \Rightarrow a = \pm b$

Conclusion $ax-by$ depends on $bx-ay$ only if $a = \pm b$

Freshman42
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  • This solution is valid only if you know that the function $f$ is differentiable, which is a substantial assumption. – John Hughes Jun 29 '16 at 11:15
  • @JohnHughes : can you find please "the valid solution when $f$ is not differentiable" ? – Freshman42 Jun 29 '16 at 15:06
  • I didn't say that there was a valid solution in which $f$ is not differentiable. I said that assuming that $f$ was differentiable was not justified by the statement of the problem, and makes the whole thing easier. It's like saying "Well, $f$ must be a quadratic, so ..." --- there's just no a priori reason for believing that. – John Hughes Jun 29 '16 at 15:11