New SkillsFormulating Maths
And & or
For all & there exists
Example 3. Let n ³ 1. Prove that (1 - [x/n] )n £ e-x whenever 0 £ x £ n.
Experimentation. You might try taking the binomial expansion of the left-hand side of the inequality, but this does not prove to be very fruitful.
Let's look at the simplest case, n = 1. Then we have to prove that 1-x £ e-x whenever 0 £ x £ 1. A sketch of the graph of e-x shows that this is true; it should be easily proved by calculus.
For general n, we can take nth roots, so the inequality to be proved becomes: 1 - [x/n] £ e-x/n. This can be deduced from the inequality in the preceding paragraph, if we replace x by x/n.
Proof. We first prove that 1-y £ e-y whenever y ³ 0. Let f(y) = y+e-y. Then f¢(y) = 1-e-y ³ 0 for all y ³ 0, so f is increasing in this range. Now, f(0) = 1, so 1 £ f(y) = y+e-y for all y ³ 0. This gives the required inequality.
Now suppose that n ³ 1 and 0 £ x £ n. Putting y = x/n in the inequality obtained in the preceding paragraph gives
It would be wise to include the sketch given above with your solution.
| Design: Paul Gartside,
Content: Prof. C. Batty,