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

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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

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.

  • 2
    $\begingroup$ that's an interesting way to concretize and visualize why such a transformation (+1) $\endgroup$ – G Cab Aug 13 at 18:18

Starting from the LHS


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




which is equivalent to


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

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


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


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





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

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.