Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 15 / c

Table of contents
- Solution c

Solution

For

x > 0

, the definition of hyperbolic cosine simplifies the composite function exactly:

cosh (ln x) = \frac{e ^{l n x} + e ^{- l n x}}{2} = \frac{1}{2} (x + \frac{1}{x}) .

Let

h = x - 2

. The geometric series gives

\frac{1}{2 + h} = \frac{1}{2} (1 - \frac{h}{2} + \frac{h ^{2}}{4} - \frac{h ^{3}}{8} + O (h^{4})) .

Adding

2 + h

and dividing by two gives the requested Taylor expansion:

cosh (ln x) = \frac{5}{4} + \frac{3}{8} (x - 2) + \frac{1}{16} (x - 2)^{2} - \frac{1}{32} (x - 2)^{3} + O ((x - 2)^{4}) .

The underlying infinite geometric expansion has

∣ x - 2∣ < 2

.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version