# Simplify expression assuming b>>c using big O notation

So I found in Approximating for $a \gg b$

that I can use big O notation to emulate a>>b behavior. However, when I try that with the following equation, I get an undesired result.

Expression:

Normal[((c + 2 b) a)/((c + d) (c + b)) /. {b -> b + O[c]}]


Expected:

2a/(c+d)


Actual return:

2a/d


Where in the above, I'm trying to state that b>>c. Therefore the b in the numerator and the b in the denominator should just cancel.

Is there something I'm missing?

• This variant seems about right. Rewrite b as a function of c and some "small" epsilon, then expand in that epsilon. In:= Normal[ Series[((c + 2 b) a)/((c + d) (c + b)) /. {b -> c/eps}, {eps, 0, 0}]] Out= (2 a)/(c + d) – Daniel Lichtblau Apr 6 '19 at 14:29
• Why not ((c + 2 b) a)/((c + d) (c + b)) + O[b, Infinity] // Normal? – Carl Woll Apr 6 '19 at 14:39
• Those ways do work for this specific equation, but they don't capture the meaning of b>>c. If I use either of those methods on ((c + 2 b) a)/((c + d) (c + b) (b + e)) I get 0. Though in hindsight, I see that the method I found doesn't really capture the meaning either. – Vince McKinsey Apr 7 '19 at 4:40

I think you can just do a series expansion of b at infinity. For example:

Series[((c+2 b) a)/((c+d) (c+b)), {b, Infinity, 0}] //TeXForm


$$\frac{2 a}{c+d}+O\left(\left(\frac{1}{b}\right)^1\right)$$

Series[((c+2 b) a)/((c+d) (c+b)(b+e)), {b, Infinity, 1}] //TeXForm


$$\frac{2 a}{b (c+d)}+O\left(\left(\frac{1}{b}\right)^2\right)$$

If only c is much smaller than b, you can use Daniel's answer in the comments:

Normal[Series[((c+2 b) a)/((c+d) (c+b)(b+\[Epsilon])) /. b -> c/\[Epsilon], {\[Epsilon], 0, 2}]] /. \[Epsilon]->c/b


(2 a)/(b (c + d)) - (a c)/(b^2 (c + d))

Compare this to the above approach where everything is smaller than b:

Normal @ Series[((c+2 b) a)/((c+d) (c+b)(b+e)), {b, Infinity, 2}]


(2 a)/(b (c + d)) + (-a c - 2 a e)/(b^2 (c + d))

Big $$O$$ notation is used to write a term to first order in a selected term. Here starting with the term

$$\frac{a\left(2b+c\right)}{\left(b+c\right)\left(c+d\right)}$$

and assuming that $$b\gg c$$, the goal is to write the fraction in first order in $$\frac{b}{c}$$. By multiplying the numerator and denominator by $$1/b^2$$ and making the substitutions $$\tilde{a}\equiv a/b$$, $$\tilde{c}\equiv c/b$$, and $$\tilde{d}\equiv a/d$$ we can rewrite the original term as

$$\frac{\tilde{a}\left(2+\tilde{c}\right)}{\left(1+\tilde{c}\right)\left(\tilde{c}+\tilde{d}\right)}$$

Now assuming that $$b\gg c$$ results in $$\tilde{c}\approx 0$$, thus we can do a Taylor expansion about the point $$c=0$$ using Series.

Series[(a (2 + c))/((1 + c) (c + d)), {c, 0, 0}]


this will give the output

2a/d + O[c]^1


Simplifying the original equation when $$b\gg c$$ is not as simple as just crossing out the $$c$$ terms in the numerator and the denominator due to the presence of the additional $$c$$ dependent cross terms resulting from the multiplication in the denominator.

• Hi cphys, thank you for the explanation. I see now why my method was erroneous. Do you know of a method to accomplish b>>c? – Vince McKinsey Apr 7 '19 at 4:43