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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?

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    @guy completely agree. This question lacks details. If it were an oral exam my first answer would have been that is is true for all values of c. Then I would have continued by saying that maybe the real question was to have zero valued first order derivative. Being approximately constant around a value implies being constant on a segment containing this value.2012-04-02

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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?

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    Sure thing, glad I could help! Sorry for the mistakes :p2012-04-01
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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 $0$.

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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\,. $

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    I know about Maclaurin's. I had included that in my question but I guess I deleted it accidentally.2012-04-01
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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 $c = 2$.

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    I really don't understand why they would solve it this way, in the exam. Finding when the derivative function is zero is far quicker than figuring the Maclaurins of both functions and so on.2012-04-03
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$(1+cx)^\frac{1}{2} = 1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^{2n-1}} \binom{2n-2}{n-1}c^n x^n .$ 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). $