Calculus Problems

The first problem asks us to show that ãΠ > Πã . The analysis from my previous post actually breezes by this, but the fact is a pretty cool property of ã. To show this, first we log both sides, to obtain Π Log[ã] > ã Log[Π], or Π Log@ΠD

>

ã Log@ãD

So, we'll look at the function f(x) =

f '(x) =

f '(x) =

x , Log@xD

defined for x Ε (0, ¥).

Log@xD - x H1xL Log@xD2

Log@xD - 1 Log@xD2

So, f '(x) = 0 when Log[x] = 1, or x = ã. This implies that f(x) is decreasing on (0,ã) and increasing on (ã,¥) with a relative minimum at x = ã. We needed to show that

Π Log@ΠD

>

ã , Log@ãD

and since f(x) increases on (ã,¥) and Π > ã, the statement must be true.

What this implies is that for any Α Ε (1, ¥), Α¹ã, ãΑ > Αã . So, ã2 > 2ã , a fact that shouldn't be obvious.

The other problem is this: f is a continuous function defined on [0,2] such that f(0) = f(2) = 0. Show that there exists a Ζ Ε [1,2] so that f(Ζ) = f(Ζ-1). First, I defined a function g(x) = f(x) - f(x-1), defined on [1,2], which will make some of the calculations we look at easier. We now have to show that g(x) has a zero. So, it is sufficient to show that at one point it is positive, and at another it is negative- since g(x) is continuous, by the Internediate Value Theorem the function must have a zero. Well, we know that g(1) is either positive or negative; if it is zero then we are already done. If it is positive, then g(2) = 0 - f(1) = -f(1) must be negative, and by the IVT g(x) must have a zero. Similarly, if f(1) is negative, then g(2) is positive, and by the IVT g(x) must have a zero on [1,2]. QED. Here are some example functions of this type and their graphs:

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Calculus Problems.nb

y = x2 -2x 0.5

0.5

1.0

1.5

2.0

Out[72]=

-0.5

-1.0

1 3 y = Πã xHx- LHx-1LHx- LHx-2L 2 2 2

1 Out[73]=

0.5

1.0

1.5

2.0

1.5

2.0

-1

-2

y = SinH2ΠxL 1.0

0.5

Out[74]=

0.5

-0.5

-1.0

1.0

Calculus Problems.nb

y = LogH-x2 +2x-1L 0.7 0.6 0.5

Out[75]=

0.4 0.3 0.2 0.1

0.5

1.0 3

1.5

2.0

1.5

2.0

2

y = ãArcTanI-x +3 x -2 xM -1 0.4 0.3 0.2 Out[76]=

0.1

0.5

1.0

-0.1 -0.2 -0.3

These all have an x value Ζ such that y(Ζ) = y(Ζ-1), which you can probably spot visually. Graphically, if I took a horizontal segment of length 1 unit and moved it around the graph, I am guaranteed a positioning where both ends of the segment are touching the graph.

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