Find $a$, $b$ and $c$ in $\frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3$

Question

Find the values of the positive constants $a$, $b$ and $c$ given that when $x$ is sufficiently small for terms in $x^4$, and higher powers of $x$, to be neglected then: $$ \frac{e^{ax}}{2+bx}=\frac{1}{2}+\frac{x^2}{4}-cx^3 \space\space\text{(assume $|bx| < 2$)} $$

Expanding $e^{ax}$ first, we get:

$$ 1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6} $$

and expanding $(2+bx)^{-1}$ we get:

$$ \frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16} $$

I tried dividing the expressions and comparing the coefficients, but didn't really get anywhere. Is there an easier way to do this?

Answer 1

Hint: Don't divide, Multiply!

$$\frac{e^{ax}}{2+bx}=e^{ax}(2+bx)^{-1}=\left(1+ax+\frac{a^2x^2}{2}+\frac{a^3x^3}{6}+\cdots\right)\left(\frac{1}{2}-\frac{bx}{4}+\frac{b^2x^2}{8}-\frac{b^3x^3}{16}\cdots\right)$$

Can you take it from here?

Answer 2

Hint

As Integrator said, multiply.

Another simple way : consider the function $$e^{a x}=(2+b x)(\frac{1}{2}+\frac{x^2}{4}-cx^3)$$ Use the Taylor series for the lhs (just as you did) and expand the rhs. Now, identify as many terms as you can. You will then get the following equations $$ a-\frac{b}{2}=0$$ $$a^2-1=0$$ $$\frac{a^3}{6}-\frac{b}{4}+2 c=0$$