Given that $\frac{a}{1-x}+\frac{b}{1+2x}\approx-3+12x$ for small values of $x$, find the value of $a$ and $b$

Question

Given that $\frac{a}{1-x}+\frac{b}{1+2x}\approx-3+12x$ for small values of $x$, find the value of $a$ and $b$.

This question is very hard i am on the chapter of Binomial expansion of $(1+x)^n$ for values of n that are not positive integers i know how to solve these questions but have no idea how to solve this type of question when involves 2 unknown variables.

Answer 1

Using $\displaystyle\frac{1}{1+y}\approx 1 - y$ for small $y$, we wish to find $a$ and $b$ such that:

$$\frac{a}{1-x} + \frac{b}{1 + 2x} \approx 12x - 3$$

$$a(1 + x) + b(1 - 2x)= 12x - 3$$

$$(a - 2b)x + (a + b)= 12x - 3$$

Now, to make all of the terms match we have the system of linear equations:

$$\begin{cases}a - 2b = 12\\a + b = -3\end{cases}$$

Solving, we get $\boxed{(a, b) = (2, -5).}$

Answer 2

As a (fill the gap) re the other responses:

$(1-x)(1 + x + x^2 + \cdots + x^n) = 1-x^{(n+1)} \approx 1$, for small values of $x$. In fact, when $|x| < 1$, the infinite series $(1 + x + x^2 + \cdots)$ converges to $\frac{1}{1-x}.$

Further, for small values of $x$, it is reasonable to discard all but the first two terms of $(1 + x + x^2 + \cdots).$

This implies that for small values of $x$, $\frac{1}{1-x}$ can be approximated by $(1 + x).$

Similarly, $(1 + 2y)\times [1 - 2y + 4y^2 - 8y^3 + \cdots - \cdots + (-1)^n (2y)^n] = 1 + [-2y]^{(n+1)}.$

Therefore, for $|y| < \frac{1}{2}$, the infinite series
$[1 - 2y + 4y^2 - 8y^3 + \cdots - \cdots ]$ converges to $\frac{1}{1 + 2y}$.

Thus, similar to the analysis in the first part of this answer, for small values of $y,~ \frac{1}{1 + 2y}$ may be approximated by $(1 - 2y).$

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The above analysis begs the question: how do you know what approximations to take.

Answer
Look at the original problem.
Since the RHS is a first degree polynomial, a blind guess is that you want to find first degree polynomials to approximate the LHS, for small values of $x$.

Unfortunately, this blind guess requires the abstract vision that such a LHS approximation will result in two (solvable) linear equations in two unknowns for the variables $a,b.$

It is reasonable to surmise that the premise that $x$ is a small value is authorizing polynomial approximations to the fractions. It is then reasonable to question what degree polynomials should be used for the LHS approximations.

You are then supposed to question as follows: how could the LHS polynomial approximations be of degree higher than $1$, and still be equivalent to the RHS polynomial?