Part 2

New Skills

Formulating Maths
Introduction
Hypotheses &C
Implications
And & or
For all & there exists
Dependence

Proof
Introduction
Counter examples
Constructing proofs
Understanding
Experimentation
Precision
Examples
Assumptions
Contradiction
Induction
The End

 

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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

1 - x n £ e-x/n.

Since x £ n, the left-hand side of this inequality is non-negative. We may therefore raise both sides to the power n, to obtain:

æ ç è 1 - x n ö ÷ ø n   £ e-x,

as required.

It would be wise to include the sketch given above with your solution.