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$$ \frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) $$ It is easy to check it by computing right hand side. It feels unnatural and a little magical. I could't derive it starting from LHS

This identity is used in the proof of Stolz Cesaro theorem (https://ru.wikipedia.org/wiki/Теорема_Штольца). It is in russian I understood with the help of google translate .

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  • $\begingroup$ If you know what you are aiming at then you can start by adding and subtracting $\frac{c-xd}{b}$ ... $\endgroup$ – Martin R Aug 13 at 16:09
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    $\begingroup$ There is a reason why the LHS is transformed like this. $b$ and $d$ do not fall from the sky. It seems you have to read the proof again more carefully. $\endgroup$ – callculus Aug 13 at 16:21
  • $\begingroup$ Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS. $\endgroup$ – steven gregory Aug 13 at 17:32
  • $\begingroup$ @callculus It is transformed like that so we can show it is less than epsilon. In the proof i have the identity does fall from the sky. If you have a more intuitive explenation please share . $\endgroup$ – Milan Aug 13 at 18:13
  • $\begingroup$ Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained. $\endgroup$ – Somos Aug 13 at 18:50
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Consider the task of cooking up a straight line by combining two other straight lines.

Given a straight line $y=-x+a$
You want to represent this as a sum of two other straight lines.
Letting the slope of one line to $-m$ forces the slope of other line to $-(1-m)$.
You can get that by comparing like terms in : $$\color{red}{-x+a} = \color{blue}{-mx + c} - m'x+c'$$ Also constant term is $a-c$.
So above equation becomes $$\color{red}{-x+a} = \color{blue}{-mx + c} + (1-m)\left(\dfrac{a-c}{1-m}-x\right)$$

enter image description here

Replacing $m$ with $\frac{d}{b}$ and $a$ with $\frac{a}{b}$ gives your identity.

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    $\begingroup$ that's an interesting way to concretize and visualize why such a transformation (+1) $\endgroup$ – G Cab Aug 13 at 18:18
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Starting from the LHS

$$\frac{a}{b}-x$$

we can subtract and add $\frac{c-xd}{b}$ to form

$$\Big(\frac{a}{b}-x\Big) - \frac{c-xd}{b} + \frac{c-xd}{b}$$

which rearranges to

$$\frac{c-xd}{b}+\Big(\frac{a}{b}-x-\frac{c-xd}{b}\Big)$$

or

$$\frac{c-xd}{b}+\Big(\frac{a-c+xd}{b}-x\Big)$$

which is equivalent to

$$\frac{c-xd}{b}+\Big(\frac{a-bx-c+dx}{b}\Big)$$

where $\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right) = \Big(\frac{a-bx-c+dx}{b}\Big)$ and therefore we have the RHS

$$\frac{c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right)$$

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Let $$ Z=\:\frac{\left(c-xd\right)}{b}+\left(\frac{b-d}{b}\right)\:\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$$ Let $$ K=\frac{\left(c-xd\right)}{b}$$ Let $$ N=\left(\frac{b-d}{b}\right)\:\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$$

$$Z=K+N$$

We will rewrite $\left(\frac{a-c}{b-d}\:-\:x\:\right)\:$ as:

$$\left(\frac{a-c}{b-d}\:-\:x\:\right)\:=\left(\frac{a-c}{b-d}\:-\:x\:\frac{\left(b-d\right)}{\left(b-d\right)}\:\right)\:$$

$$N\:=\left(\frac{b-d}{b}\right)\left(\frac{a-c}{b-d}\:-\:x\:\frac{\left(b-d\right)}{\left(b-d\right)}\:\right)\:=\left(\frac{b-d}{b}\right)\left(\frac{a-c\:-x\left(b-d\right)}{b-d}\:\right)$$

$$N\:=\left(\frac{a-c\:-x\left(b-d\right)}{b}\:\right)$$

$$N\:=\left(\frac{a-c\:-xb+xd}{b}\:\right)$$

Using $Z=K+N$ and substituting the last expression for $N$ we get:

$$z\:=\frac{\left(c-xd\right)}{b}+\left(\frac{a-c\:-xb+xd}{b}\:\right) $$

$$z\:=\left(\frac{a\:-xb}{b}\:\right)$$

$$ z\:=\frac{a\:}{b}-x$$

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