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I have the following equation: $a(x)k-b(x)=0\Rightarrow k=\dfrac{b(x)}{a(x)}$. I find approximate solutions around $x=0$ by two ways:

(1) I expand the equation as $[a(0)+xa'(0)]k-[b(0)+xb'(0)]=0\Rightarrow k=\dfrac{b(0)+xb'(0)}{a(0)+xa'(0)}$.

(2)I expand the solution itself, $k=\dfrac{b(x)}{a(x)}\approx \dfrac{b(0)}{a(0)}+x\left(\dfrac{\partial(b/a)}{\partial x}\right)_{x=0}.$

Why solution (2) is better than solution (1)? And in what cases both the solutions can be almost equal?

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1 Answer 1

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Both solutions are valid for small $x$, but (2) is better because it is simpler while having the same magnitude of error with respect to the original solution $b(x)/a(x)$. The way we express this is to say that both approximations are valid to $O \left ( x^2 \right )$. Google "big-O notation" for more details.

You can see that both approximations are equivalent to this magnitude of error by Taylor expanding approximation (1) and writing out the derivative term in approximation (2).

Let $a(0) := a$ and $b(0) := b$, etc. Then

$$ \frac{a + a' x}{b + b' x} = \frac{a}{b} \frac{1+ \frac{a'}{a} x}{1+ \frac{b'}{b}x}$$

$$ = \frac{a}{b} \left [ 1 + \left ( \frac{a'}{a} - \frac{b'}{b} \right ) x + O \left ( x^2 \right ) \right ] $$

$$ = \frac{a}{b} + \left ( \frac{a'}{b} - \frac{a b'}{b^2} \right ) x + O \left ( x^2 \right ) $$

Note that the term in the parentheses in the last expression is the derivative of $a(x)/b(x)$ at $x=0$.

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