# Quod Erat Demonstrandum

## 2008/09/29

### Just a question of applied math. from a F.7 boy

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 6:13 下午
Tags: , ,

Just discuss an easy AL Applied Mathematics (II) Question with students.

Let $f(x), g(x), h(x)$ be twice differentiable functions such that $f(x) = g^2(x) + x^3h(x)$.

(a) Let $p(x) = \frac{f(x)}{g(x)}$, where $g(0) \ne 0$. Show that $p(0) = g(0), p'(0) = g'(0), p''(0) = g''(0)$.

(b) Using (a), or otherwise, find Taylor’s expansion of the function $\frac{2x^4 - 3x^3 + x + 4}{\sqrt{x + 4}}$ about $x = 0$, up to the term in $x^2$.

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In answering (a), $p(x)$ can be written as

$p(x) \equiv g(x) + x^3u(x)$ where $u(x) = \frac{h(x)}{g(x)}$

Hence $p(0) = g(0)$

The usual practice, differentiate w.r.t. $x$, obtaining

$p'(x) \equiv g'(x) + x^3u'(x) + 3x^2u(x)$, then $p'(0) = g'(0)$

$p''(x) \equiv g''(x) + x^3u''(x) + 6x^2u'(x) + 6xu(x)$, then $p''(0) = g''(0)$

A student, Fu, queried that when the term $x^3u''(x)$ is a constant (i.e. $u''(x)$ cancel $x^3$), the step of putting $x = 0$ into $x^3u''(x)$ may not be zero! Good question!

Friends, is it necessary to worry about it?