Help with Taylor series for proving the approximation of a function

Question

Recently I got the following question:

If $a$ and $b$ are nearly equal, show that $$\left(\frac{5a-2b}{3b}\right)^{\frac13} \approx \frac a{a+2b} + \frac{b+a}{3b}.$$

I have attempted to solve this using a Taylor series:
$$ \frac{5a-2b}{3b} = x\\ \text{Taylor series around $x=1$:}\\ f(1) = 1\\ f'(1) = \frac13\\ f(x) \approx 1 + \frac13(x-1)+...\\ \text{Substituting }\frac{5a-2b}{3b} = x,\\ f(a) \approx \frac{5a+4b}{9b} \approx \frac{b+2a}{9b}+\frac{b+a}{3b} $$
This is where I have reached so far, my lecturer hasn't covered or even talked about Taylor series and I have tried to understand it on my own. I think the last part might have a problem, although I'm not sure exactly.

I also think this could be done using binomial expansion, but I'm not sure how to go around it.

Any help or advice is appreciated, thank you in advance!

Answer 1

Your derivation looks correct, what remains to show is that for $a\approx b$ we have $\frac a{a+2b}\approx \frac{b+2a}{9b}$.

What do we mean by $\approx$ for two functions? In terms of comparing two functions: It depends on the context, but here one might argue that $f(x)\approx g(x)$ near $x=x_0$ means that $f(x_0) = g(x_0)$ and $f'(x_0) = g'(x_0)$ (i.e. same value and same tangent). One would write that $f(x) - g(x) = O((x-x_0)^2)$, which means that the corresponding Taylor series match on the first 2 terms, but might differ on the higher-order terms.

You wouldn't say that $f(x)\approx g(x)$ if $f'(x_0) \neq g'(x_0)$. In this case, the Taylor series of the functions would only match on the first (zero order) term.

Application to your expressions. To make this rigorous, write $a = b + \epsilon$, with $\epsilon$ "small" (to express $a\approx b$). We take the difference of the two terms we need to compare, and substitute $a$: $$\frac a{a+2b}- \frac{b+2a}{9b} = \frac{-2\epsilon^2}{9b(3b+\epsilon)} = \frac{-2}{27b^2}\epsilon^2 + O(\epsilon^3).\qquad (*)$$ You can justify the last identity by using the Taylor expansion of $1/(3b+\epsilon)$ around $\epsilon = 0$: $$\frac 1{3b+\epsilon} = \frac1{3b} - \frac1{9b^2}\epsilon +O(\epsilon^2).$$ When we multiply this by $\frac{-2\epsilon^2}{9b}$, we find (again, order by powers of $\epsilon$): $$\frac{-2\epsilon^2}{9b(3b+\epsilon)} = \frac{-2}{27b^2}\epsilon^2 + \frac{2}{81b^3}\epsilon^3 + (\epsilon^4) = \frac{-2}{27b^2}\epsilon^2 + O(\epsilon^3).$$ The $O(\epsilon^3)$-notation in $(*)$ encapsulates any remaining higher-order terms of the form $c_3\epsilon^3 + c_4\epsilon^4 +\cdots$, containing no smaller powers of $\epsilon$.

This is as good as it can get if you use the first order Taylor expansion. "To first order", your solution is just as valid as the expression in the problem statement. If you would want higher accuracy, you would need more terms in the expansion of $\sqrt[3]{\cdot}$.

See https://en.wikipedia.org/wiki/Big_O_notation for further explanation on the notation.

Answer 2

Let $a=k b$ which makes the lhs to be $$\sqrt[3]{\frac{1}{3} (5 k-2) }$$ Use the binomial expansion or Taylor series around $k=1$ $$\sqrt[3]{\frac{1}{3} (5 k-2) }=1+\frac{5 (k-1)}{9}-\frac{25}{81} (k-1)^2+O\left((k-1)^3\right)$$

The rhs write $$\frac{k+1}{3}+\frac{k}{k+2}=1+\frac{5 (k-1)}{9}-\frac{2}{27} (k-1)^2+O\left((k-1)^3\right)$$

They are not the same but $$\text{lhs - rhs}=-\frac{19}{81} (k-1)^2+O\left((k-1)^3\right)$$ which is not much since $k\approx 1$