# Clustering around the rainbow ray

We want to show that for any point the interval coming from any small interval centred on is larger than the interval coming from the interval of the same width as , but centred on (see the figure).

Figure 7: The interval J1 is smaller than the interval J2.

Since is the only point with we know that To match with our figure we’ll assume that An analogous argument works when

Now choose a positive number with (you’ll see later why we make this choice). From the definition of a derivative we know that there exists such that for all with we have

 (1)

Similarly, there exists such that for all with we have

 (2)

Now choose any less than both and . Let be the interval and let be the interval

Since is the only value for which we know that the maximum of on must occur at one of the end-points of . Call that point . This means that the width of the interval is

Since is the only value for which we also know that the slope of is negative throughout the interval . On the function achieves its maximum at the lower end point of and its minimum at the upper end point The width of the interval is therefore

Now since both and lie within of , we know from inequality 2 that

and

(The inequalities have changed direction because, since , we have multiplied by a negative number in each case.)

Putting this together gives

 (3)

Since lies within of , we know from inequality 1 that

 (4)

And since we have chosen so that , putting inequalities 3 and 4 together gives

Thus, interval is wider than interval . QED.

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