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Kartik's question at Yahoo! Answers regarding differentiation of a continued fraction

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Kartik,

We are given:

\(\displaystyle y=\cfrac{x}{a+\cfrac{x}{b+\cfrac{x}{a+\cfrac{x}{b+ \cdots}}}}\)

and asked to find \(\displaystyle \frac{dy}{dx}\) in terms of $y$ only.

We may choose to write:

\(\displaystyle y=\cfrac{x}{a+\cfrac{x}{b+y}}\)

Multiplying through by \(\displaystyle \frac{1}{b+y}\) we obtain:

\(\displaystyle \frac{y}{b+y}=\frac{x}{a(b+y)+x}\)

Inverting both sides, then subtracting through by 1, we have:

\(\displaystyle \frac{b}{y}=\frac{a(b+y)}{x}\)

Solving for $x$, we obtain:

\(\displaystyle x=ay+\frac{a}{b}y^2\)

Differentiating with respect to $y$, we find:

\(\displaystyle \frac{dx}{dy}=a+\frac{2a}{b}y=\frac{a(b+2y)}{b}\)

Hence:

\(\displaystyle \frac{dy}{dx}=\frac{b}{a(b+2y)}\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
... inverting both sides, then subtracting through by 1, we have...

\(\displaystyle \frac{b}{y}=\frac{a(b+y)}{x}\)

Solving for $x$, we obtain...
If You want to obtain $\displaystyle \frac{d y}{d x}$ in 'standard form' [i.e. as function of the only x...] You can solve respect to y obtaining...


$\displaystyle y= - \frac{b}{2} \pm \sqrt{\frac{b^{2}}{4} + \frac{b}{a} x}\ (1)$

... and then differentiate (1). Note from (1) that y(x) is a multivalue function...

Kind regards

$\chi$ $\sigma$