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.

———————————————-

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?

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