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I've been working through old MIT practice exam papers, and I found a question that stumped me. It goes:
What value for the constant $c$ will make the function $e^{-x}\sqrt{1+cx}$ approximately constant, for values of $x$ near $0$? (Show your work.)
The answer begins with saying that $e^{-x}\sqrt{1+cx} \approx (1-x)(1+\frac{cx}{2})$
I'm not sure where this came from.
Can someone explain to me how this is solved?
It's the first two terms of each power series for $e^{-x}$ and $\sqrt{1 + cx}$.
We want it to be "approximately constant" so looking at the approximation
$$f(x) = e^{-x}\sqrt{1+cx} \approx (1-x)\left(1+\frac{cx}{2}\right) = 1 + x\left(\frac{c}{2} - 1\right) - \frac{cx^2}{2}.$$
We need to figure out some way to make this constant near $0$. Do you know something about derivatives that will be able to help you figure that out?
$$f'(x) = \frac{c}{2} - 1 - cx.$$
We're looking near $x = 0$, and want $f$ to be constant. What does that mean in terms of the derivative?
The idea is to replace each of the factors by its Maclaurin series,
$$e^{-x}\sqrt{1+cx}=\left(\sum_{n\ge 0}(-1)^n\frac{x^n}{n!}\right)\left(\sum_{n\ge 0}\binom{1/2}n(cx)^n\right)\;,$$
and then approximate this by ignoring the non-linear terms in the two series:
$$e^{-x}\sqrt{1+cx}\approx (1-x)\left(1+\frac12cx\right)\;.\tag{1}$$
Note: The binomial coefficient $\binom{r}n$ is probably unfamiliar when $r$ is not a non-negative integer; it’s just an abbreviation for $$\frac{r(r-1)(r-2)\dots(r-n+1)}{n!}\;,$$ so $\binom{r}0=1$, and $\binom{r}1=r$.
Multiply out the righthand side of $(1)$:
$$e^{-x}\sqrt{1+cx}\approx 1+\left(\frac{c}2-1\right)x-\frac12cx^2\;.$$
You want this to be approximately constant when $|x|$ is small. For $x$ near $0$, which changes faster, $x$, or $x^2$? It’s $x$, right? The graph of $y=x^2$ is practically horizontal near $0$, while the graph of $y=x$ has slope $1$. Thus, you want to choose $c$ to kill off the $x$ term, so that $$e^{-x}\sqrt{1+cx}\approx 1-\frac12cx^2\;,$$ and the function is nearly constant at $1$ for $x$ near $0$. This evidently means taking $c=2$, and we have finally $$e^{-x}\sqrt{1+2x}\approx 1-x^2\approx 1$$ for $x$ near $0$.
There's a way that's almost as quick if not quicker, but it doesn't use the hint. A function is constant if its derivative is $0$, so just take the derivative, set it equal to $0$, factor a bit and one arrives at the equation $\frac12 c - 1 - cx = 0$. Now if $x=0$, then $c = 2$.
You should read about Taylor series expansions of functions. It's a way to approximate functions at a point by a power series, or if you truncate, by a polynomial. If this is new to you, don't worry about a few underlying complications with making these expansions, just learn how to generate them for simple functions. For example, $$ \exp(-x) \approx 1 - x + \tfrac{1}{2}x^2 - \tfrac{1}{6}x^3\,. $$
$(1+cx)^\frac{1}{2} = 1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^{2n-1}} \binom{2n-2}{n-1}c^n x^n .$ Hence $$(1+cx)^\frac{1}{2} e^{-x} = (1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^{2n-1}} \binom{2n-2}{n-1}c^n x^n)(\sum_{m=0}^\infty \frac{(-1)^m x^m}{m!})\\ =1+ \sum_{n=1}^\infty (-1)^{n-1} (\sum_{k=0}^n \frac{1}{k2^{2k-1}} \binom{2k-2}{k-1}c^k \frac{1}{(n-k)!} ) x^n\\ =1+ (\frac{c}{2} -1)x+ (\frac{1}{2} - \frac{c}{2} - \frac{c^2}{8})x^2 +O(x^3)$$
So, if the linear term is zero(i.e., $c=2$), it is $$1 - \frac{1}{2} x^2 +O(x^3). $$
