# How to derive $\frac ab -x= \frac {c-xd}{b}+\left(\frac{b-d}{b}\right) \left(\frac{a-c}{b-d} - x \right)$

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

• If you know what you are aiming at then you can start by adding and subtracting $\frac{c-xd}{b}$ ... – Martin R Aug 13 at 16:09
• 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. – callculus Aug 13 at 16:21
• Step by step reduce the LHS to the RHS. Turn that upside down and you derive the RHS to the LHS. – steven gregory Aug 13 at 17:32
• @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 . – Milan Aug 13 at 18:13
• Do you have a reference for the proof? If so, please include the link in your question in order to make it self contained. – 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 $$\color{red}{-x+a} = \color{blue}{-mx + c} + (1-m)\left(\dfrac{a-c}{1-m}-x\right)$$

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

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

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

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

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:

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

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

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