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?

  • 7
    $\begingroup$ It's the first two terms of each power series for $e^{-x}$ and $\sqrt{1 + cx}$ $\endgroup$
    – Tyler
    Apr 1, 2012 at 20:40
  • $\begingroup$ a.k.a. Taylor (or Maclaurin) series (since $x_0=0$). So then your function $\approx 1+O(x^2)$ for $c=2$. $\endgroup$
    – bgins
    Apr 1, 2012 at 20:43
  • $\begingroup$ I thought it was Maclaurin's. Thanks :) $\endgroup$ Apr 1, 2012 at 20:46
  • 1
    $\begingroup$ I interpreted this to mean that the linear approximation to $e^{-x} \sqrt{1 + cx}$ should be a flat line at $0$, i.e. $f'(0) = 0$ where $f(x) = e^{-x} \sqrt{1 + cx}$. So why mess around with Taylor series (presumably there IS a reason, I just don't know it)? Also, isn't there some ambiguity in saying that the function should be "approximately constant" near a certain value? If a function is continuous at $x$ then it is "approximately constant" near $x$. $\endgroup$
    – guy
    Apr 1, 2012 at 21:29
  • $\begingroup$ @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. $\endgroup$ Apr 2, 2012 at 10:20

5 Answers 5


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?

  • $\begingroup$ If the function is to be "approximately constant", you want to get rid of the linear term, not the quadratic term. $\endgroup$
    – Ted
    Apr 1, 2012 at 20:47
  • $\begingroup$ Whoops, misread the question! I fixed it. $\endgroup$
    – Tyler
    Apr 1, 2012 at 20:49
  • 2
    $\begingroup$ The way I see it, the idea is to centre the maximum of the function over $x=0$, which would mean the function would be pretty much constant, and I can do that by killing off the term $x(\frac{c}{2}-1)$. If $c=2$, that gives me what I want. Thanks for helping me understand. $\endgroup$ Apr 1, 2012 at 21:20
  • $\begingroup$ Sure thing, glad I could help! Sorry for the mistakes :p $\endgroup$
    – Tyler
    Apr 1, 2012 at 21:21

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 $c = 2$.

  • $\begingroup$ The hint just makes it easy to differentiate, I guess. $\endgroup$ Apr 2, 2012 at 7:08
  • $\begingroup$ I'm not so sure the approach suggested by the hint is easier. The product rule is pretty quick for $e^{−x}\sqrt{1+cx}$, and you don't have to remember the Taylor series nor decide how many terms to include. But you're right that a second degree polynomial is really easy to differentiate. $\endgroup$
    – Patrick
    Apr 2, 2012 at 18:37
  • $\begingroup$ This seems like a smarter approach. $\endgroup$ Apr 2, 2012 at 22:34
  • $\begingroup$ 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. $\endgroup$ Apr 3, 2012 at 15:31

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, $$ \exp(-x) \approx 1 - x + \tfrac{1}{2}x^2 - \tfrac{1}{6}x^3\,. $$

  • $\begingroup$ I know about Maclaurin's. I had included that in my question but I guess I deleted it accidentally. $\endgroup$ Apr 1, 2012 at 20:57

$(1+cx)^\frac{1}{2} = 1 + \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n2^{2n-1}} \binom{2n-2}{n-1}c^n x^n .$ 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 $$1 - \frac{1}{2} x^2 +O(x^3). $$


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