Year: 1998
Paper: 3
Question Number: 7
Course: UFM Pure
Section: Taylor series
No solution available for this problem.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Sketch the graph of ${\rm f}(s)={ \e}^s(s-3)+3$ for $0\le s < \infty$. Taking
${\e\approx 2.7}$, find the smallest positive integer, $m$, such that
${\rm f}(m) > 0$.
Now let
$$
{\rm b}(x) = {x^3 \over \e^{x/T} -1} \,
$$
where $T$ is a positive constant.
Show that ${\rm b}(x)$ has a single turning
point in $0 < x < \infty$. By considering the behaviour for small $x$
and for large $x$, sketch ${\rm b}(x)$ for $0\le x < \infty$.
Let
$$
\int_0^\infty {\rm b}(x)\,\d x =B,
$$
which may be assumed to be finite.
Show that $B = K T^n$ where $K$ is a constant, and $n$ is an
integer which you should determine.
Given that $\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}$,
use your graph of ${\rm b}(x)$ to find a rough estimate for $K$.